J50I 


ELEMENTARY 

Reinforced  Concrete  Building  Design 

by 

LEONARD   C.  URQUHART 
Instructor  of  Bridge  Engineering,  Cornell  University 


ELEMENTARY 

Reinforced  Concrete  Building  Design 

•» 

by 

LEONARD   C.   URQUHART 
Instructor  of  Bridge  Engineering,  Cornell  University 


CARPENTER  &   COMPANY 

ITHACA.  NEW  YORK 

1915 


v) 


COPYRIGHT 
1915 

CARPENTER  &  Co. 


INTRODUCTION 

The  following  pages  are  intended  to  furnish  the  computing  side  of 
an  elementary  course  in  reinforced  concrete  construction  and  to 
supplement  a  text  book  used  in  the  class  room.  The  references 
made  to  Diagrams  and  Tables  are  to  those  of  Reinforced  Concrete 
Construction,  Volume  I,  by  George  A.  Hool. 

It  has  been  the  intention  to  give  a  complete  design  for  the  roofs, 
floors,  columns,  and  footings  of  a  reinforced  concrete  building  of 
the  beam-and-girder  type.  Other  details  such  as  walls,  windows, 
stairways,  elevator  shafts,  etc.,  do  not  as  a  rule  affect  the  main 
design,  involve  more  practical  considerations,  and  will  not  be 
taken  up  here.  For  an  example  of  these  details  the  student  is 
referred  to  Volume  II  of  the  text  book  mentioned  above. 

The  design  of  the  floors  and  columns  is  given  in  a  very  precise 
form,  and  although  so  much  exactness  is  not  required  in  ordinary 
building  design,  it  has  been  the  intention  to  familiarize  the  student 
with  methods  applicable  to  large  and  important  members,  and  also 
to  show  methods  of  checking  each  portion  of  the  design. 

There  are  many  methods  of  form  construction,  but  the  one 
briefly  given  in  the  following  design  seems  to  be  more  general  than 
many  others.  Here  again  more  exactness  will  be  found  than  is 
ordinarily  required,  but  in  buildings  having  many  similar  units 
such  exactness  would  result  in  final  economy. 

The  design  of  a  combined  footing  is  given  in  order  to  furnish  an 
outline  for  a  problem  of  this  type  in  the  computing  room. 

CORNELL  UNIVERSITY, 
October  31,   1915. 


SPECIFICATIONS 

1 .  The  concrete  in  this  structure  shall  consist  of  a  wet  mixture 
of  one  part  Portland  cement  to  six  parts  of  aggregate,  fine  and 
coarse,  and  shall  be  capable  of  developing  an  average  compressive 
stress  of  2000  pounds  per  square  inch  at  twenty-eight  days  when 
tested  in  cylinders  eight  inches  in  diameter  and  sixteen  inches  long 
under  laboratory  conditions  of  manufacture. 

2 .  All  reinforcement  shall  consist  of  medium  open  hearth  steel 
having  an  elastic  limit  of  not  less  than  50000  pounds  per  square 
inch,  and  shall  be  of  plain  round  section. 

3.  All  flexure  formulae  shall  be  based  on  the  "straight  line 
theory." 

4.  The  span  lengths  used  in  computations  shall  be  taken  as  the 
distance  center  to  center  of  supports. 

5.  Slabs,  beams,  and  girders  shall  be  considered  as  fully  con- 
tinuous,  and  bending  moments  shall  be  computed  accordingly 
both  for  uniform  and  concentrated  loads. 

6.  The  negative  moment  which  exists  at  the  supports  shall  be 
taken  equal  to  the  maximum  positive  moment  sustained  by  the 
slab,  beam,  or  girder. 

Reinforcement  shall  be  provided  to  resist  this  negative  moment 
and  all  such  reinforcement  shall  be  continued  beyond  the  support 
to  the  quarter  point  of  the  adjacent  span,  but  in  no  case  shall  this 
distance  be  less  than  that  required  to  develop  the  full  strength  of 
the  rods  at  the  allowable  bond  stress. 

7.  The  lateral  spacing  of  parallel  rods  shall  not  be  less  than  two 
and  one-half  diameters  center  to  center,  nor  shall  the  distance  from 
the  side  of  the  beam  or  girder  to  the  center  of  the  nearest  rod  be 
less  than  two  diameters.     Layers  of  rods,  in  beams  and  girders ? 
shall  have  at  least  one-half  inch  vertically  in  the  clear.     In  girders 
and  columns  the  metal  shall  be  protected  by  a  mininium  of  two 
inches  of  concrete ;    (i.  e.  from  the  surface  of  the  rod  to  the  surface 
of  the  concrete)  in  beams,  by  one  and  one-half  inches;    in  floor 
slabs,  by  one-half  inch  of  concrete.     Main  reinforcement  in  slabs, 
beams,  and  girders  shall  not  be  less  than  three-eighths  of  an  inch 
nor  more  than  one  inch  in  diameter.     Stirrups  shall  not  be  less 
than  one  quarter  inch  in  diameter. 


8 .  In  columns  the  concrete  to  a  depth  of  one  and  one-half  inches 
shall  be  considered  as  protective  covering  and  not  included  in  the 
effective  section. 

9.  The  minimum  allowable  thickness  for  slabs  is  four  inches, 
and  the  main  reinforcement  when  consisting  of  rods  shall  not  be 
placed  farther  apart  than  two-and-one-half  times  the  slab  thick- 
ness. 

Rods  placed  perpendicular  to  the  main  reinforcement  for  the 
prevention  of  cracks,  due  to  shrinkage,  etc.,  shall  be  provided  to  an 
amount  not  less  than  one-fourth  of  one  percent  of  the  slab  section. 

10.  Beams  whose  axes  are  perpendicular  to  the  main  reinforce- 
ment of  the  slab  may  be  designed  as  Tee  beams,  provided  the  slab 
is  poured  with  the  beams  and  properly  bonded  to  the  same.     Under 
these  conditions  the  effective  width  of  flange  shall  be  taken  so  as 
not  to  exceed    (a)  one-fourth  of  the  span  length  of  the  beam,(b) 
eight   times   the   thickness  of   the   slab   plus   the   width   of  the 
stem. 

11.  When  the  maximum  shearing  stresses  exceed  the  value 
allowed  for  the  concrete  alone,  web  reinforcement  must  be  provided 
to  aid  in  carrying  the  diagonal  tension  stresses. 

The  horizontal  reinforcement  is  to  be  bent  up  where  possible 
to  assist  in  taking  the  diagonal  tension  stresses,  but  this  is  not  to 
be  considered  in  determining  the  required  amount  of  web  reinforce- 
ment. 

In  the  calculations  of  web  reinforcement  the  concrete  may  be 
counted  upon  as  carrying  one-third  of  the  shear,  and  the  remainder 
is  to  be  provided  for  by  means  of  metal  reinforcement. 

The  maximum  longitudinal  spacing  of  vertical  stirrups  shall  not 
exceed  three-fourths  of  the  working  depth  of  the  beam.  The 
effective  length  of  stirrups  in  bond  shall  be  taken  as  six-tenths  of 
the  working  depth  of  the  beam. 

12.  The  ratio  of  length  to  diameter  of  concrete  columns  shall 
not  exceed  fifteen. 

Columns  shall  not  have  less  than  one  per  cent,  nor  more  than  four 
per  cent,  of  longitudinal  reinforcement,  which  shall  be  held 
together  by  bands  or  hoops,  not  less  than  one-fourth  inch  in 
diameter,  spaced  center  to  center  not  more  than  one-half  of  the 
diameter  of  the  column. 


Not  less  than  four  rods  shall  be  used  for  the  reinforcement  of  any 
column,  and  rods  shall  not  be  less  than  one-half  inch  nor  more  than 
one  inch  in  diameter. 

Bending  stresses  due  to  eccentric  loads  must  be  provided  for  by 
increasing  the  section  until  the  maximum  stress  does  not  exceed  the 
allowable  value. 

13 .  The  footings  shall  be  so  designed  that  they  can  be  poured  at 
one  operation.     The  bottom  of  the  footing  shall  not  be  nearer  to 
the  lowest  layer  of  steel  than  four  inches. 

14.  The  allowable  working  stresses  shall  not  exceed  the  follow- 
ing: 

Axial  compression  on  columns  reinforced  with  longitudinal  steel 

and  spirals  or  bands 500  pounds  per  square  inch 

Concrete  in  compression  in  slabs,  beams,  or  girders 

650  pounds  per  square  inch 

Concrete  in  compression  over  the  support  of  continuous  slabs, 

beams  and  girders 750  pounds  per  square  inch 

Shearing  strength  of  plain  concrete  as  a  measure  of  diagonal  ten- 
sion           40  pounds  per  square  inch 

Shearing  strength  of  concrete  with  proper  metal  reinforcement 

120  pounds  per  square  inch 

Bond  stress  between  concrete  and  plain  reinforcing  rods  v 

80  pounds  per  square  inch 

Tensile  stress  in  reinforcing  steel ....  16000  pounds  per  square  inch 
Compressive  stress  in  reinforcing  steel  not  more  than  fifteen 
times  the  working  compressive  stress  in  the  concrete. 
Ratio  of  the  modulus  of  elasticity  of  steel  to  that  of  concrete ....  15 


The  building  whose  partial  design  is  given  in  the  following  pages 
has  three  stories  above  ground  and  a  basement,  and  has  the  follow- 
ing dimensions: 

Width,  center  to  center  of  wall  columns 54/o" 

Length,  center  to  center  of  wall  columns loo'o* 

Span  of  slabs 9'  o" 

Span  of  beams 20'  o* 

Span  of  girders 18'  o* 

Distance  floor  to  floor 12'  o" 

It  is  designed  for  a  live  load  on  the  floors  of  100  pounds  per  square 
foot,  a  live  load  on  the  roof  of  35  pounds  per  square  foot,  and  the 
allowable  soil  pressure  is  taken  as  4000  pounds  per  square  foot. 


DESIGN    OF   THE    FLOOR    SLAB 

Assumed  thickness  from  Table  6 4  inches 

Total  load  per  square  foot 150  pounds 

Bending  moment  for  unit  slab  ^  x  1 50  x  92  x  1 2  =  1 2 1 50  in-lbs. 

From  Table  2p= 0077 

k  = 368 

j  =    : .    .874 

107.4 

12150  . 

=     3.07  inches 

107.4  x  12 

Taking  the  value  of  d  to  the  nearest  quarter  of  an  inch  greater 
than  the  theoretical  value  determined,  and  adding  either  three 
quarters  of  one  inch  or  one  inch  for  insulation,  so  that  the  total 
thickness  of  the  slab  is  determined  to  the  nearest  one-half  inch,  the 
actual  value  of  d  is  3><  inches  and  the  total  thickness  of  the  slab  4 
inches  as  assumed. 

12150  .     , 

-    267  square  inches 


i6ooox  .874x3.25 

150x4.5  . 

e0  =  o  -  Q  -  =  ....................   2.97  inches 


Using  |  inch  round  rods  s  =  x  12  =  ....   4.  96  inches 

1.178 

or  s     =  —  —  x  12  =  .....   4.  76  inches 
2.97 

Rods   will   be   spaced  ............  4  J   inches   center   to   center. 

To  provide  for  the  negative  bending  moment  near  the  support 
each  alternate  rod  is  bent  upward  near  the  quarter  point  of  the 
span  and  continued  over  the  support  to  the  quarter  point  of  the 
adjacent  span.  This  distance  beyond  the  support  must  not  be 
less  than  that  required  to  develop  the  full  strength  of  the  rod  in 
bond  which  is  asfs  -?-  perimeter  of  rod  x  u'  for  a  f  inch  rod  the 
distance  is  i8f  inches  which  is  less  than  one-quarter  of  the  span 
(27  inches). 

The  reinforcement  required  in  the  other  direction  is  .0025  x  4  x  12 
=  .12  square  inches  per  foot  section  of  slab,  requiring  that  f  inch 

round  rods  be  placed  '-  -  xi2   =   11.04  or  n  inches  center  to 
.120 

center. 


IO 

REVIEW    OF   THE    FLOOR   SLAB 

.1104 
p  =  -  —     =     .................................  0075 

4.5X3-25 

From  Table  3  k=  ................................  375 

j  =    .........  -  ......................  875 

as  for  a  unit  slab  =  -  :  x  .1104  ..........  294  square  inches 

4-5 

t        12150 

L  =  -  -  -    =    ..........................     145500 


=  2x  14500  x.  0075  =  8o 

•375 

T^        12150 

K=  -  ^=2    =     ..............................        95.9 

12x3.25 

From  Diagram  i,  with  the  value  of  K  =  95.9  and  p  =  .0075,  fs  = 
14500  and  fc  =  580  which  check  the  above  computations. 

150x4.5 
v  =  -  ^r-   ^  -  =  ............................    19.8 

i2X.875X3.25 

1  50x4.5 

u  =  -  -  -  —  -  =  76.1; 

12  x  i.i78x.875X3-25 


*   '-REINFORCEMENT 
$"<j>4-f'c,TOC. 


-8-0" 


I  EACH   ALTERNATE   ROP  LIKE  a 

ROP  a 


I.  rrA.'of,&7-     rr     T&  .- J.   rr  .1 


ROPbr 


FlGURE     I 


DESIGN    AND    REVIEW    OF   THE    FLOOR    BEAM 

The  weight  of  slab  carried  by  one  floor  beam  is  50  x 

20  x  9  =   ..................................     9000  pounds 

The  live  load  carried  by  one  floor  beam  is  100x20x9=  1  8000  pounds 
The  weight  of  the  beam  below  the  slab  may  be  assumed 

as  10%  of  the  above  weight  =  .................     2700  pounds 

29700  pounds 


II 


The  bending  moment  is  TV  x  29700  x  20  x  12  =  594000  inch  pounds 
In  order  that  the  allowable  unit  shearing^stf^SF  shall  not  be 

exceeded,  the  beam  must  have  a  cross-section  b'd  equal  to  -* — - 

IOS 

=  141.4  square  inches.     With  b'  =  7!  inches,  d  =  18.25  inches. 
For  economy  with  b'  =  7  p  and  the  ratio  of  cost  between  the  steel 


and  concrete  taken  as  60,  the  depth  d  should  be 


60  x  594000       4 

16000  X  7f  2 


=  18.95  inches.  Therefore  d  must  be  equal  to  or  greater  than 
18.25  inches  and  as  nearly  equal  to  18.95  inches  as  the  forms  will 
permit.  (See  actual  sizes  of  lumber  under  Design  of  Forms) .  The 


I  I4x4"  ROUGH 

Li    CAP 

FIGURE  2 

distance  to  be  covered  by  the  side  lagging  (see  Figure  2)  must  then 
be  at  least  18.25  +  2J  +  if  -  4  =  18.75  inches.  Two  pieces 
Ji'  x  5!"  and  one  piece  if*  x  7!*  furnish  19}  inches;  d  is  taken  as 
i8f  inches,  and  the  total  depth  of  the  beam  i8j  *  +  2p  =  i\\ 
inches.  The  distance  of  2\  inches  allows  for  two  rows  of  rods, 
whose  centers  are  i  \  inches  apart  vertically,  and  assuming  that  the 
rods  are  not  larger  than  one  inch  in  diameter,  this  allows  a  clear 
distance  below  the  lower  rods  of  i  \  inches  according  to  the  specifi- 
cations. 
The  actual  weight  of  the  beam  below  the  slab  is 

17^x7!  x  i5ox  20    = 
144 

The  revised  total  load  on  the  beam 29830  pounds 

The  revised  bending  moment 596600  inch  pounds 

The  revised  maximum  shear , 14915  pounds 

Taking  the  value  of  jd  equal  to  d  -  ft, 

596600  14915 


2830  pounds 


i6jx  16000 


=  2.23  sq.  in.  and  s0  = 


i6|x8o 


=    11.13 


m- 


12 

Six  |  inch  round  rods  furnish  an  area  of  2.65  square  inches  and  a 
total  perimeter  of  14.14  inches,  but  over  the  support  the  stresses 
in  the  steel  and  in  the  concrete  cannot  be  brought  within  the  allow- 
able value  unless  all  of  the  lower  rods  are  carried  through  to  the 
quarter  point  of  the  adjacent  span.  Five  J  inch  round  rods, 
furnishing  very  little  more  steel  do  not  require  these  long  lower 
rods  and  are  adopted.  The  latter  furnish  an  area  of  3.01  square 
inches  and  a  total  perimeter  of  13.75  inches. 

The  width  of  the  flange  (6)  is  taken  as  large  as  the  specifications 
will  permit,  i.  e.  either  J  x  20'  -o"  =  60  inches,  or  8  x  4  -f  yf  = 
39.75  inches;  the  latter  value  (40  inches)  governing  in  this  case. 

p  =  -  =  .0040  and  -5  =  -33  =    .21.     On    Diagram    9 

18.75  x  40  d       18} 

the  point  located  by  the  above  values  of  p  and  -.  falls  above  the 
dotted  line  showing  that  the  neutral  axis  is  in  the  web. 
From  Table  9,  with-  =  .21  andp  =  .0040 

k  =  .304  /  =  .914 

From  Diagram  9,  k  =  .30  and  /  =  .91,  thus  checking  the  above 
values  from  Table  9. 


f  s  =  -  —  -—  =   ........................     11600 

3.01  x  .9i4x  18.75 

.314  x  11600 

fc  =  15  d  -  .304)   =      ............................ 

Three  rods  from  each  beam  are  bent  up  and  carried  over  the 
support  to  the  quarter  point  of  the  span  of  the  adjacent  beam. 
One  of  these  rods  from  each  beam  reaches  its  upper  position 
directly  over  the  center  of  the  support  so  that  these  two  rods 
together  only  have  the  effective  strength  of  one  rod.  Thus  with 
three  rods  from  each  beam  bent  up  the  effective  steel  area  over  the 
support  is  equal  to  five  f  inch  round  rods  or  3.01  square  inches. 
The  two  rods  from  each  beam  which  are  not  bent  up  are  carried 
beyond  the.  ends  of  the  rods  from  the  adjacent  beam  far  enough  to 
develop  the  full  strength  of  the  rods  in  bond,  (43.75  inches  for  a 
J  inch  rod)  making  the  effective  steel  area  at  the  bottom  of  the 
beam  over  the  support  equal  to  two  j  inch  round  rods  or  1.20 
square  inches.  In  the  case  of  the  six  f  inch  round  rods  mentioned 


13 

above  it  would  have  been  necessary  to  carry  three  rods  from  each 
beam  through  along  the  bottom  to  the  quarter  point  of  the  adjacent 
span  in  order  to  obtain  a  large  enough  steel  area  to  keep  the  unit 
stresses  within  the  allowable  limits. 

(For  beams  having  an  even  number  of  rods,  one-half  of  the  total 
number  of  rods  from  each  beam  can  be  bent  up  and  carried  through 
to  the  quarter  point  of  the  adjacent  beam,  while  the  remaining  half 
is  carried  past  the  rods  from  the  adjacent  beam  far  enough  to 
develop  the  full  strength  of  the  rods  in  bond.  This  gives  an 
effective  area  at  the  top  of  the  beam  equal  to  the  total  number  of 


FULL  LINES  GIVE 
LENGTHS  OF  ROPS 
FOR  FLOOR  BEAM; 
POTTEP   LINES 
FOR    ROOF 
BEAM. 


FIGURE  3 

rods  in  one  beam,  and  an  effective  area  at  the  bottom  equal  to  one- 
half  of  this  amount.) 

The  points  at  which  the  three  rods  may  be  bent  up  are  deter- 
mined as  follows : 

One  rod  at  —  I  i  -    /  f     1  =  5- 53  feet  from  the  support. 
One  rod  at  ^  f  i  -    /  _     J  =  3  •  68  feet  from  the  support. 

One  rod  at  2-  fi  -   /  3     1  =  2 . 26  feet  from  the  support. 

These  points  are  also  checked  graphically  by  means  of  Figure  3. 


The  upper  layer  of  rods  over  the  support  is  placed  as  high  as  the 
slab  rods  will  permit  so  that  their  centers  may  be  considered  to  be 
i|  inches  from  the  upper  surface  of  the  slab. 

b  =  7!  inches 

d  =  21  J-  ij-f 19! inches 

d'  =  2f  inches 

as  =  3.01  sq.  in. 

i. 20  sq.  in. 

3.01 


a'. 


P  = 

P'  = 
P'  = 
P 

d'  = 
d  = 


.0204 
.0082 
•4 

•143 


(See  Figure  4) 
From  Table  1 1,  interpolating  f or  p  =  .0204 

p'  =  -25P  P'  =  -5op  p'  =  -4P 

L        K        j         L        K        j         L        K 
d' 
d 

d' 


d 

d' 

d 


10     .262     .0173     .848    .305    .0177    .864     Interpolating 
15     .255     .0169    .836    .288    .0172    .845 


143  .256     .0170    .838    .291    .0173    .848    .277    .0172    .844 

596600 

-=   749 


I  2  100 


*•*  •"*    ^T! 
•I    i 

!    1     J 


STIRR 


I       I 


itthemidspanis.9i6x  i8f  =  17.18 
The  value  of  yd  at  the  support  is.  844  x  19  J  =  16.25 


FIGURE  4 


Ji  The  value  of  jd  would  of  course  change  at  the 
£  I  point  of  inflection  but  the  design  will  be  on  the 
£_[  side  of  safety  if  the  smaller  value  of  jd  computed 

above  is  used  in  all  of  the  computations  involving 

shear  and  bond. 


15 

The  maximum  shear  at  the  support  is 149*5  pounds 

The  maximum  shear  at  midspan  is  J  x  i 8000  =   .  .     2250  pounds 

The  unit  shear  at  the  support  is     3149*5    -   =    ..     118.5  pounds 

?4   X    l0-25 

The  unit  shear  at  midspan  is  — g —  -    = 17.9    pounds 

74  x  10.25 

The  distance  from  the  support  beyond  which  stirrups  are  not 
required  (i.  e.  the  point  where  the  unit  shear  is  equal  to  40  pounds 

WlX2 

per  square  inch)  may  be  determined  by  the  equation  — ^  +  wdx  = 
4objd  -f-     — ,  where  x  =  the  distance  from  the  section  to  the 

opposite  support,  Wd  =  the  dead  load  per  linear  foot  and  wi  =  the 

live  load  per  linear  foot. 

wj  =  100  x  9  =  900  pounds,  wa  =  50  x  9+  142  =  592  pounds 

9oox2  .  3       .  592  x  20  , 

-  -f  592X  =  40  x  7!  x  16.25  +  — »  x  +  26-32  x  =  487-0. 

2  X  2O  2 

and  x  =  12.53  feet>  so  that  stirrups  are  required  over  a  distance  of 
20  -  12.53  =  7.47  feet  from  the  support.     Selecting  a  stirrup  that 

requires  a  spacing  of  about  6  inches  at  the  support,  as  =  -    X 

149 1 5  x —   _  square  inches;    a  f  inch  round  U  stirrup 

16000  x  16.25 

furnishes  .2208  square  inches  and  is  adopted. 

WlX2  1 

The  maximum  shear  at  any  section  is  — =-  +  wa  (x  -  -  ). 


W, 

Wdl 

->  =  22.5 

—  =  5920 

Dis.  from 
Support 

x 

x2 

WlX2 
~2T 

WdX 

V 

3 

Spacing 
inches 

0 

20 

400 

9000 

11840 

14920 

86100 

5-8 

i 

19 

36i 

8120 

11250 

13450 

86100 

6-4 

2 

18 

324 

7290 

10660 

12030 

86100 

7.2 

3 

17 

289 

6500 

10060 

10640 

86100 

8.1 

4 

16 

256 

5760 

9470 

9310 

86100 

9-2 

5 

15 

225 

5060 

8880 

8020 

86100 

10.7 

6 

H 

196 

4410 

8290 

6780 

86100 

12.7 

7 

13 

169 

3800 

7700 

5580 

86100 

154 

7-47 

12.53 

157 

3530 

7420 

5030 

86100 

17.1 

i6 

The  maximum  spacing  allowed  by  the  specifications  is  f  x  i8f  = 
14.06  inches  so  that  in  drawing  a  curve  for  the  stirrup  spacings 
similar  to  Figure  46  of  the  text,  it  becomes  a  horizontal  line  when 
its  ordinate  reaches  a  value  of  14  inches.  In  drawing  this  curve  it 
is  necessary  that  the  horizontal  and  vertical  scales  be  the  same  in 
order  that  the  spacings  plotted  as  ordinates  may  be  transferred  to 
the  horizontal  axis  by  means  of  45°  lines.  These  lines  cannot  be 
drawn  until  the  location  of  the  first  stirrup  is  determined,  which 
will  be  placed  two  or  three  inches  from  the  edge  of  the  support,  and 
the  width  of  the  column  or  girder  which  supports  the  floor  beam  is 
not  yet  known. 

„,     ,       ,  j      i  ^u     -L-  •         .2208  x  16000 

The  bond  stress  developed  on  the  stirrups  is — 

2  x  1.178  x  .6  x  18.75 

=  133  pounds  per  square  inch.     This  high  value  is  safe,  however, 

since  the  upper  ends  of  the  stirrups  are  bent  into  the  form  of  hooks. 

The  bond  stress  on  the  upper  horizontal  rods  at  the  support  is 


14915 


5  x  2.74QX  16.25 
allowable  value 


=  66.8  pounds  per  square  inch  and  is  within  the 


FIGURE  5 

DESIGN    AND    REVIEW    OF    GIRDER. 

The  maximum  moment  in  the  girder  has  the  same  value  when  all 
of  the  load  in  one  bay  is  considered  as  uniformly  distributed  on  the 
girder,  as  it  does  when  the  load  from  one  floor  beam  is  considered 
as  concentrated  at  the  center  of  the  girder.  The  total  load  sup- 
ported by  two  floor  beams  is  59660  pounds,  and  the  weight  of  the 
girder  below  the  slab  may  be  assumed  as  i  itLS,  where  L  is  the  span 
of  the  beam  and  5  the  span  of  the  slab.  This  assumed  weight  is 
11x4^x20x9  =  8910  pounds,  and  the  total  load  is  68570  pounds. 


1234260  inch  pounds 


M  =  —  x  68570  xi8xi2  = 
12 

MI  =  1,234,260  = 

107.4 
With  6=13^  inches,  d  must  be  at  least  29.2  inches. 

The  distance  to  be  covered  by  the  side  lagging  must  be  at  least 
29.2  -f-  2\  +  if  -  4  =  29.45  inches  (see  Figure  6).  Four  pieces 
7f  x  1  1  furnish  31  inches;  d  is  taken  as  30.75  inches  and  the  total 
depth  of  the  beam  (h)  as  33.25  inches. 


'^~ 

^Tl 

.          1 

. 

ft, 

JHT* 


4x4"  CAP 


i* 


I.I 

FIGURE  6 
The  actual  weight  of  the  beam  below  the  slab  is 


144 


x  150 


x  18  =  7410  pounds,  the  revised  total  load  67070  pounds,  and  the 
revised  bending  moment  1207260  inch  pounds. 

1207260 

as  =  -r-     —^ =  2. 8 1  square  inches. 

16000  x  .874x30.75 

Five  |  inch  round  rods  furnish  an  area  of  3.01  square  inches  and 
can  be  placed  in  one  row  as  assumed. 


P  =  7TT 


3.01 


13^x30.75 

Table  3,  k= 

j  =    


1,207,260 


K  = 


3.01  x. 877x30.75 

2  X  .OO72  X  I49OO 

•369 
I,2O7,26o 


I3-5X  30-75 


.0072 

•369 

.877 

14900 

582 
94.6 


i8 


From  Diagram  i  with  K  =  94.6  and  p  =  .0072,  f,  =  15000  and  f 
=  580,  which  check  the  above  computations. 

In  order  more  nearly  to  approach  the  actual  conditions  of  load 
distribution  in  the  girder,  it  is  assumed  that  the  girder  supports  the 
slab  and  its  live  load  for  a  distance  on  each  side  of  the  girder  equal 
to  its  own  width,  and  the  load  carried  by  one  floor  beam  not  in- 
cluded in  the  above  is  assumed  as  concentrated  at  the  center  of 
the  girder.  Accordingly, 


(i)  The  uniform  load  = 


^JLIll 


lg  _|_ 


=  ^520  pounds 


(2)  The  concentrated  load  =  29830  -  9110  -r-  2  =   25275  pounds 
Mu  =  /j  x  16520  x  i8x  12  =   .........    297360  inch  pounds 

Mc  =  1x25275x18x12  ............   909900  inch  pounds 


FIGURE  7 

Using  these  two  moments  Figure  7  is  constructed  and  the  points 
at  which  the  rods  may  be  bent  up  are  determined.  Both  Figures 
3  and  7  neglect  the  continuity  of  the  construction  and  therefore 
give  very  conservative  results. 

As  both  the  bending  moment  and  the  cross-section  of  the  beam 
have  the  same  value  at  the  support  as  at  midspan,  five  rods  will  be 
required  at  the  top  of  the  beam  over  the  support,  while  none  is 
required  at  the  bottom.  However,  it  is  better  design  to  carry  the 
lower  rods  through  in  a  manner  similar  to  that  of  the  floor  beam. 

25275  +  16520 


Maximum  shear  at  the  support 


=    20900  pounds 


19 

2  r  2  7  C 

Maximum  shear  at  midspan  -      -  =    12640  pounds 

Unit  shear  at  the  support —  =  .  .        57.4  pounds 

I3-5X30-75X.877 

Unit  shear  at  midspan —  =  .  14.7  pounds 

13-5  x  30.75  x. 877 

The  true  maximum  shear  at  midspan  would  occur  when  a  strip 
equal  to  three  times  the  width  of  the  girder  times  the  span  of  the 
slab  is  unloaded,  which  would  give  a  value  of  13020  pounds,  but  the 
difference  is  so  small  that  it  may  be  neglected  and  the  shear  may  be 
assumed  to  vary  uniformly  throughout  the  span. 

The  point  beyond  which  stirrups  are  not  required  is  — 

=  2.10  feet  from  midspan.     Selecting  a  stirrup  which  requires  a 

4  x  20000  x  6 
spacing  of  about  6  inches  at  the  support  as  =  — - — 

16000  x  30.75  x  .877 

=   .194  square  inches;    a  finch  round  U  stirrup  furnishes  .2208 
square  inches  and  is  adopted. 

™  .  .  .   f  x  .2208  x  16000 

Ine  spacing  required  at  the  support  is ; =    6.8 

57-4  x  13} 

inches  and  that  at  a  point  2.10  feet  from  midspan  9.8  inches.  A 
curve  similar  to  that  used  for  the  floor  beam  may  be  constructed 
by  connecting  the  points  located  by  the  two  spacings  computed 
above  by  a  straight  line,  and  the  actual  spacing  of  the  stirrups 
determined  when  the  width  of  the  column  is  known. 
The  actual  bond  stress  on  the  horizontal  rods  is 

20900  .  , 

=  56.4  pounds  per  square  inch. 

5  X2.749X  .877  x  30.75 

A  rod  schedule  similar  to  Figure  5  can  now  be  constructed  for 
the  girder. 

DESIGN    OF    THE    ROOF. 

The  design  of  the  roof  is  similar  to  that  of  the  floor,  but  as  the 
loads  are  smaller,  the  construction  will  be  somewhat  lighter  and 
only  the  design  will  be  considered,  the  construction  not  being  heavy 
enough  to  call  for  a  review. 

Roof  Slab.  Assuming  the  minimum  allowable  thickness  of  slab 
which  is  four  inches,  the  total  load  carried  is  50  +  35  =  85  pounds 


20 

per  square  foot.  The  bending  moment  for  a  unit  slab  is  6885  inch 
pounds  and  the  required  depth  to  the  steel  2.31  inches.  Using  a 
depth  as  large  as  the  thickness  of  the  slab  will  permit  (i.  e.  3.25 
inches),  the  steel  area  required  per  unit  slab  is  .152  square  inches, 
and  f  inch  round  rods  spaced  8  J  inches  center  to  center  furnish  this 
area.  The  reinforcement  required  in  the  other  direction  is  .12 
square  inches  per  foot  section  which  is  furnished  by  f  inch  round 
rods  spaced  n  inches  center  to  center.  The  main  reinforcing 
rods  are  bent  upward  at  the  same  points  as  those  used  in  the  floor 
slab. 

Roof  Beam.  The  total  weight  of  slab  and  live  load  sustained 
by  the  beam  is85X2ox9  =  15300,  and  assuming  the  weight  of  the 
beam  as  10%  of  this  load  the  total  assumed  load  is  16830  pounds, 
the  end  shear  8415  pounds,  and  the  bending  moment  336600  inch 
pounds.  The  required  value  of  b'd  for  shear  is  80.2  square  inches 
and  with  b'  =  5!  inches,  d  must  be  at  least  13.9  inches.  The 
economical  value  of  d  is  16.8  inches,  and  the  distance  to  be  covered 
by  the  side  lagging  should  be  somewhat  more  than  13.9  +  2f  +  if 
-4  =  14.4  inches.  Two  pieces  if"  x  yf "  furnish  15.5  inches  mak- 
ing the  value  of  d  15.0  inches  and  the  actual  weight  of  the  beam 

below  the  slab  -^ — —  x  150x20=    1650  pounds.     The  revised 
144 

values  of  the  total  load,  end  shear,  and  bending  moment  are  16950 
pounds,  8475  pounds,  and  339,000  inch  pounds  respectively. 

The  area  of  steel  required  is  1.63  square  inches  and  the  total 
perimeter  8.15  inches,  which  can  be  furnished  by  three  f  inch  round 
rods  with  an  area  of  1.80  square  inches  and  a  total  perimeter  of  8.25 
inches. 

The  points  at  which  two  of  the  rods  may  be  bent  up  can  be 
determined  by  the  use  of  Figure  3 ,  the  dotted  lines  representing  the 
length  of  rods  for  the  roof  beam. 

It  will  be  assumed  that  an  effective  area  of  steel  at  the  top  of 
the  beam  over  the  support  of  three J  inch  round  rods  and  one  J  inch 
round  rod  at  the  bottom  of  the  beam  will  be  sufficient  to  keep  the 
unit  stresses  within  the  allowable  values.  Figures  similar  to  2,  4, 
and  5  can  now  be  constructed  for  the  roof  beam. 

The  maximum  shear  at  the  center  of  the  beam  is  f  x  6300  =  788 
pounds,  and  taking  jd  =  d  -  \t  —  13  inches,  the  unit  shear  at  the 


21 

support  is  113.5  pounds  and  at  midspan  10.5  pounds.  Since  the 
live  load  is  small  it  may  be  assumed  that  the  shear  varies  uniformly. 
Then  the  point  beyond  which  stirrups  are  not  required  is 

— — —  =  2.86  feet  from  midspan  or  7.14  feet  from  the  sup- 

113-5-10.5 

port.  By  using  the  same  method  of  finding  the  location  of  this 
point  as  was  used  in  the  design  of  the  floor  beam,  this  distance  is 
found  to  be  7.03  feet  from  the  support  so  that  the  error  in  the  above 
computation  is  small  and  on  the  side  of  safety. 

Selecting  a  stirrup  that  requires  a  spacing  of  about  six  inches  at 

the  support  a^  =  fx    .      =    .163  square  inches;  a  A  inch 

16000 x 13 

round  U  stirrup  furnishes .  1 534  square  inches  and  is  adopted.  The 
spacing  required  at  the  support  is  5.6  inches  and  at  the  point  7.14 
feet  from  the  support  16.0  inches.  According  to  the  specifications 
11.25  inches  is  the  maximum  allowable  distance  between  stirrups. 
With  these  spacings  as  ordinates  a  curve  of  stirrup  spacings  can  be 
constructed  similar  to  that  of  the  floor  girder. 

Roof  Girder.  The  total  load  sustained  by  two  roof  beams  is 
33900  pounds  and  assuming  that  the  weight  of  the  girder  below  the 
slab  is  3.2LS2  pounds  =  5180  pounds,  the  total  assumed  load  is 
39080  pounds,  and  the  bending  moment  703440  inch  pounds.  Then 
bd2  =  6550,  and  with  6  =  i  if  inches,  d  must  be  at  least  23.6  inches, 
and  the  distance  to  be  covered  by  the  side  lagging  must  be  at  least 
23.6  +  2\  -f-  if  -  4  =  24.1  inches.  Three  pieces  if*  x  5!"  and 
one  piece  if*  x  7}"  furnish  25  inches,  making  d  =  24.5  inches  and 

the  actual  weight  of  the  girder  below  the  slab  — — —  x  1 50  x  18  = 

144 

5070  pounds,  and  the  revised  total  load  38970  pounds. 

The  revised  bending  moment  is  701,460  inch  pounds,  and  the 
steel  area  required  is  2.05  square  inches,  which  is  furnished  by  four 
|  inch  round  rods,  which  can  all  be  placed  in  one  row,  as  assumed. 

Assuming  the  same  distribution  of  loads  as  in  the  floor  girder, 

the  uniform  load  is—       -  x  85  x  18  +  5070  =  9570  pounds,  and 


22 

the  concentrated  load  16950  -  =  14700  pounds,  the  bending 

moment  due  to  the  uniform  load  172,260  inch  pounds,  and  that 
due  to  the  concentrated  load  529,200  inch  pounds.  With  these 
moments  a  diagram  similar  to  Figure  7  is  constructed  and  it  is 
found  that  two  of  the  rods  may  be  bent  up  at  distances  of  6.45  feet 
and  4.05  feet  respectively  from  the  support. 

The  maximum  shear  at  the  support  is  12135  pounds  and  the  unit 
shear  48.2  pounds,  and  the  corresponding  values  at  midspan  are 
7350  pounds  and  29.2  pounds  respectively.  The  point  beyond 
which  stirrups  are  not  required  is  3.36  feet  from  the  support.  A 
T5^  inch  round  U  stirrup  requires  a  spacing  of  6.5  inches  at  the  sup- 
port and  a  spacing  of  7.8  inches  at  a  point  3.36  feet  from  the  sup- 
port. The  difference  between  these  spacings  is  so  slight  that  no 
diagram  need  be  constructed  and  the  stirrups  may  be  spaced  7 
inches  center  to  center  over  the  distance  they  are  required.  Fig- 
ures similar  to  5  and  6  can  now  be  constructed  for  the  roof  girder. 

COLUMNS. 

Two  columns  are  to  be  designed:  one,  the  typical  interior 
column  supporting  the  main  floor,  and  being  in  turn  supported  by 
the  basement  footing;  two,  an  exterior  column  or  wall  column 
supporting  the  main  floor  and  having  a  girder  framing  into  it  whose 
axis  is  perpendicular  to  the  wall.  In  low  buildings  the  columns  are 
often  made  the  same  size  throughout,  but  in  high  buildings  large 
quantities  of  material  may  be  saved  by  decreasing  the  sizes  of  the 
columns  toward  the  top  of  the  building.  The  same  general  method 
of  design  applies. 

The  eccentric  moment  in  the  column  due  to  a  possible  eccentric 
load  on  the  floor  is  to  be  taken  equal  to  one-fourth  of  the  moment 
produced  by  the  eccentric  load  in  the  girder  framing  into  the 
column.  This  reduction  is  allowable  on  account  of  the  stiffness 
and  the  continuity  of  the  construction. 


REQUIREP    WORKING    SIDE  OF  A   SQUARE 
COLUMN  FOR   KNOWN  LOAPS  ANP  ECCENTRICITIES 


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12345 
ECCENTRICITIES    IN   INCHES 

FIGURE  8 

DESIGN    OF    INTERIOR    COLUMN 

Weight  of  roof  supported  by  one  column 38970  pounds 

Weight  of  floor  supported  by  one  column 67070  pounds 

Weight  of  three  floors  and  roof  supported  by  one 

column 240180  pounds 

Live  load  on  one  floor  beam    18000  pounds 

The  moment  produced  in  the  girder  due  to  this  live  load  on  the 
floor  beam  is  648000  inch  pounds,  and  one-fourth  of  this  moment 

M 

is  162000  inch  pounds.     The  approximate  value  of  x    =       ,    = 


24 

-  =  .67  inches,  and  referring  to  Figure  8  the  probable  size  of 
240100 

column  is  found  to  be  23  "  x  23  ".     With  one  and  one-half  inches  of 
fireproofing  on  all  sides  the  outside  dimensions  of  the  column  are 

2  f)  X    26 

26"  x  26".     The  weight  of  one  column  is  —  '•    —  x  150  x  12  =  8450 

144 

pounds,  and  of  three  columns  25350  pounds. 

The  total  load  to  be  supported  by  the  basement  column  with  the 
main  floor  loaded  on  one  side  of  the  column  only  is  240180  +  25350 
—  18000  =  247530  pounds.  x0  =  .65  and  referring  again  to  Figure 
8,  the  column  selected  is  of  sufficient  size. 

As  it  is  uneconomical  to  use  steel  reinforcement  in  columns,  the 
smallest  amount  allowable  by  the  specifications  is  used.  One 
per  cent.  of23X23  =  5.29  square  inches,  which  can  be  furnished  by 
twelve  f  inch  round  rods,  the  combined  area  of  which  is  5.30  square 
inches. 

a  +  nas  =  23  x  23  +  15  x  5.30  =  ...................       609 


s  =  5.3ox(8.i7)2  =    .............................        354 

=  23320+15x354=    ..........................    28630 

247530       162000x11.5 
-  ' 


And  checking  this  result  by  another  method 

12  x  .4418 
Po  =  -  =    ................................  oioo 

23x23 

r      8.17      ' 

l=^=  .......................................  35S 

f2  =  .............................................  126 

z  =  .i26x  180  =  .................................      22.7 

K=  —  —  -  -  +  '^X  —  r  -  -  -  =.870  +  .138=1.008 
i  +  15  x.  oioo        23       i  +22.  7  x.  oioo 

23x23  4/ 


25 

Using  Diagram  13,  K  =  i.oo  and  fc  =  468. 

Under  full  load,  with  no  eccentricity  to  consider,  the  stress  in  the 
concrete  only  reaches  436  pounds  per  square  inch.  According  to 
the  specifications  the  rods  in  the  column  are  held  together  by  one- 
quarter  inch  bands  placed  about  nj  inches  center  to  center. 

DESIGN    OF    EXTERIOR    COLUMN. 

The  weight  of  roof  supported  by  one  column  is: 

Live  load  180  square  feet  at  35  pounds 6300  pounds 

Slab  load  180  square  feet  at  50  pounds 9000  pounds 

Weight  of  i \  beams  below  the  slab  i  J  x  1650 2470  pounds 

Weight  of  \  girder  below  the  slab  \  x  5070 2540  pounds 


20310  pounds 

In  a  similar  manner  the  weight  from  one  floor  is  34950  pounds. 

Assuming  that  a  six  inch  wall  is  to  fill  the  space  between  columns 
and  beams,  the  weight  on  each  beam  due  to  the  wall  above  it  is 
(12  -  f  |)  (20  -  f  |)  (T6f)  150  =  13610  pounds.  As  this  is  less  than 
one-half  of  the  live  and  slab  loads  carried  by  one  floor  beam  (28080 
pounds)  if  the  wall  beam  is  made  equal  in  section  to  the  inter- 
mediate beam  it  will  sustain  the  wall  in  addition  to  its  proportion  of 
floor  load.  The  weight  of  walls,  floors,  and  roof  is  then  165990 
pounds,  and  the  total  load  on  one  floor  beam  being  29830  pounds, 
the  eccentric  moment  is  i  ,073 ,880  inch  pounds  and  the  approximate 
value  of  x0  is  1.62  inches.  From  Figure  8  the  probable  size  of  the 
column  is  22"  x  22"  and  with  ij  inches  of  fireproofing  on  all  sides 
the  outside  dimensions  of  the  column  are  2  5  *  x  2  5  *,  and  the  weight 
of  one  column  7810  pounds.  The  total  load  is  189420  pounds,  and 
the  actual  value  of  x0  is  1.42  inches,  and  referring  again  to  Figure  8 
the  size  of  the  column  selected  is  sufficient.  One  per  cent,  of  2 2  X2  2 
=  4.84  square  inches,  and  as  eight  f  inch  round  rods  furnish  only 
4.81  square  inches,  twleve  J  inch  round  rods  furnish  the  smallest 
section  allowed  by  the  specifications. 

ac  -f  nas  =  564,  Ic  =  19520,  r  =  7.78,  Is  =  321,  I  =  24330,  and 
fc  =  458- 

r  r2 

p0  =  .0109,  -  =  .353,-  =  .125,  z  =  22.5,  K  =  1.170,  andfc    = 

458- 


26 

Considering  the  slight  difference  in  the  size  of  the  interior  and 
exterior  column  it  would  be  advisable  to  make  them  of  the  same 
size  for  the  sake  of  the  uniformity  of  construction. 

COLUMN    FOOTINGS. 

In  the  design  under  consideration  it  is  assumed  that  no  property 
line  interefers  with  the  construction  of  the  footings  and  a  simple 
square  footing  is  to  be  designed  for  each  column. 

Recent  tests  have  shown  that  it  makes  little  or  no  difference 
whether  rods  in  a  footing  are  placed  in  two  or  in  four  directions. 
The  design  of  the  two  way  type  follows.  For  an  example  of  the 
four  way  footing  see  Concrete  Plain  and  Reinforced  by  Taylor  and 
Thompson  or  Reinforced  Concrete  Construction  Vol  II  by  George  A. 
Hool. 

Tests  have  also  shown  that  a  diagonal  tension  failure  will  not 
occur  nearer  the  edge  of  the  column  than  at  a  distance  equal  to  the 
depth  of  the  footing  at  the  edge  of  the  column.  In  most  footings 
where  this  condition  is  considered  no  web  reinforcement  is  required 
and  in  case  the  unit  shear  in  the  concrete  does  slightly  exceed  the 
allowable  value  at  the  section  mentioned  above,  a  slight  increase  in 
the  depth  of  the  footing  will  bring  the  unit  shear  within  the  limiting 
value.  For  a  design  of  a  footing  with  stirrups  see  Concrete  Plain 
and  Reinforced. 

DESIGN    OF    INTERIOR    COLUMN    FOOTING 

Load  from  Roof '.   38970  pounds 

Load  from  three  floors 201210  pounds 

Load  from  four  columns 33800  pounds 

Bearing  area  required  2  68.50  sq.  ft. 

4000 

Allowing  about  10%  for  the  weight  of  the  footing  and  taking  the 
length  of  the  side  of  the  footing  to  the  nearest  one-half  foot,  a  foot- 
ing 9'  -  o"  x  9'  -  o"  furnishing  an  area  of  81  square  feet  is  adopted. 


8l      23X23 
Referring  to  Figure  9  the  area  of  the  trapezoid  abed  is  144 


=  1  9  -3  3  square  feet.  The  upward  pressure  of  the  soil  on  this  area 
is  19.33  x  4000  =  77300  pounds.  The  distance  of  the  center  of 
gravity  of  the  trapezoid  from  the  edge  of  the  column  (ab)  is 


BANDS  i  TC.TO  c 


i  "PIPE: 

PLATES 


w>^ 


r 


-2-9" 


. 


i\  '  i 
i/i  1 1 


3-6" 


« 2-9"- 


,-,'f>    6z"c.TOC. 


IN  THE  CLEAR 


LJJj   '   l_Hl.L 

Sl-Ui+if 

I-I--I- 


42^  /23  +  2  x  io8\ 
T  V     23  +  108    ) 


3     \^     23  +  108 
the  edge  of  the  column  is  1998000  inch  pounds. 


FIGURE  9 
25.85  inches,  and  the  bending  moment  at 


Depth  required  for  moment  .  /— ^ =  28.4  inches 

^107.4x23 


28 

Depth  required  for  punching  shear =32.0  inches 

1008000 

as   =  — ^-r = 4.46  square  inches 

16000  x. 875x32 

2o  =        773°° = 34.5inches 

Sox. 875x32 

The  effective  reinforcement  is  to  be  spaced  over  a  distance  equal 
to  the  width  of  the  column  plus  twice  the  depth  of  the  footing  at 
the  edge  of  the  column  plus  one-half  of  the  remaining  distance  = 

23  + 2x32  H =97^  inches;  J  inch  round  rods  spaced  6|  inches 

center  to  center  furnish  in  a  width  of  97  J  inches  an  area  of  6.63 
square  inches  and  a  total  perimeter  of  35.3  inches. 

The  top  of  the  footing  is  made  42  inches  square,  thus  providing 
a  6  inch  ledge  all  around  the  outside  of  the  column  so  that  the 
column  forms  may  be  erected  before  the  basement  floor  is  poured. 
The  depth  T  is  taken  equal  to  about  one-half  of  d  or  16  inches. 
Then  the  depth  of  the  footing  at  a  distance  of  32  inches  from  the 

working  side  of  the  column  (along  ef)  is  16  H -x   16  =    21.10 

oo 

inches,  and  the  distance  ef  is  108  -  2  x  loj  =  87  inches.  The  area 
of  the  trapezoid  cdef  is  7.11  square  feet,  making  the  upward  pres- 
sure from  the  soil  on  this  area  28440  pounds  and  the  unit  shear 

along  ef  equal  to —   =  17.7  pounds. 

87  X2i.i  x  .875 

The  total  maximum  depth  of  the  footing  at  the  column  edge  is 
made  37  inches  and  at  the  outside  edge  21  inches,  thus  allowing 
4  inches  clear  insulation  below  the  lower  layer  of  rods. 

The  weight  of  this  footing  is  about  31000  pounds,  making  the 
total  load  sustained  about  305000  pounds,  requiring  an  area  of 
76.13  square  feet,  or  a  footing  8.72  feet  square.  Therefore  no 
revision  is  necessary. 

DESIGN    OF    EXTERIOR    COLUMN    FOOTING 

Load  from  roof   .' 203 10  pounds 

Load  from  three  floors 104850  pounds 

Load  from  three  walls 40830  pounds 

Load  from  four  columns 31240  pounds 


197230  pounds 
Bearing  area  required 49.3 1   square  feet 


29 

Proceeding  as  in  the  design  of  the  interior  column  footing,  a 
footing  7'-6*x7'-6*  is  selected.  The  area  of  the  trapezoid 
abed  is  13.22  square  feet,  the  upward  pressure  of  the  soil  on  this 
area  52900  pounds,  and  the  bending  moment  at  the  edge  of  the 
column  934,000  inch  pounds.  The  depth  required  for  moment  is 
19.9  inches,  and  that  for  punching  shear  22.9  inches.  Adopting  a 
depth  of  23  inches  the  steel  area  required  is  2.90  square  inches 
and  the  total  perimeter  32.8  inches.  The  effective  reinforcement 
will  be  spaced  over  a  distance  of  79  inches,  and  J  inch  round 
rods  spaced  5^  inches  center  to  center  will  furnish  in  a  width  of 
79  inches  an  area  of  6.35  square  inches  and  a  total  perimeter  of  33.8 
inches. 

Making  the  top  of  the  footing  42  inches  square  and  the  depth  T, 
12  inches,  the  depth  of  the  footing  at  a  distance  of  23  inches  from 
the  working  side  of  the  column  is  17.04  inches  and. the  distance  ef 
68  inches.  The  area  of  the  trapezoid  cdef  is  6.03  square  feet,  the 
upward  pressure  from  the  soil  on  this  area  24120  pounds  and  the 
unit  shear  along  6/23.8  pounds.  The  total  maximum  depth  of  the 
footing  at  the  column  edge  is  made  28  inches  and  at  the  outside 
edge  17  inches. 

FORMS. 

i .  The  forms  are  to  be  like  those  shown  on  posted  sheet,  and  in 
accordance  with  the  following  specifications : 

Material  to  be  of  spruce,  and  the  allowable  unit  stresses  as  follows 

Modulus  of  elasticity 1300000  pounds  per  square  inch 

Extreme  fibre  stress  in  flexure  .  1 200  pounds  per  square  inch 
Compression  on  sides  of  fibres  300  pounds  per  square  inch 

Compression  on  ends  of  fibres  .  1200  pounds  per  square  inch 
Shear  parallel  to  the  grain  ....  80  pounds  per  square  inch 

Compression  on  struts  of  length 

under  1 5  times  least  diameter 

of  side  900  pounds  per  square  inch 

Compression  on  struts  over  1 5 

times  least  dimension  d.     1 200 

(i  - 1  /6od) pounds  per  square  inch 

Deflection  not  to  exceed  span  /  360. 

Basing  the  deflection  on  the  values  specified  above  with  E  = 
1,300,000  and  D  =  Span  /  360,  the  following  equations  result  and 
are  to  be  used : 


30 
Simple  beams.     Equation  i ,  W  =    277 ,000  I  /I2 


Equation  2,1  = 

Partially  continuous  beams. 
Equation  3,  W  =    462,000  I /I2 
Equation  4, 1  =     / 462,000  I/w 

S/ 

On  the  basis  of  strength 

Simple  beams.     Equation  5,  W  =  —. — ': 

i7       4.-      *  1          /g6oo.I 
Equation  6, 1  = 

Partially  continuous  beams 

Equation  7,  W  =  12000.! 

~1 e 


/  277,000  I/w 


T^  o    1  /  I2OOO.I 

Equation  8, 1  = 


In  the  above  formulae  W  =  total  load  in  pounds  on  the  span; 
w  =  the  load  per  linear  inch  of  span;  /  =  the  length  of  the  unsup- 
ported span  in  inches ;  I  =  the  moment  of  inertia  of  the  board  or 
plank  in  biquadratic  inches. 

The  forms  shall  be  designed  for  a  construction  live  load  of  75 
pounds  per  square  foot,  in  addition  to  the  weight  of  the  concrete 
they  sustain.  The  weight  of  the  forms  may  be  neglected. 

In  the  design  of  the  column  forms,  the  yokes  are  to  be  spaced  on 
the  assumption  that  the  wet  concrete  exerts  a  fluid  pressure  of  80 
pounds  per  square  foot  on  the  lagging,  but  the  yokes  in  no  case  are 
to  be  spaced  further  apart  than  24  inches  center  to  center. 

The  thickness  of  lumber  for  the  forms  depends  upon  the  number 
of  times  it  is  to  be  used  and  the  manner  in  which  it  is  erected,  but 
the  general  practice  is  i  inch  stock  dressed  to  J  inch  for  floors,  i  J  or 
i \  inch  stock  dressed  to  i  J  or  if  for  columns  and  the  sides  of  beams 
and  girders,  and  2  inch  stock  dressed  to  if  inches  for  beam  and 
girder  bottoms. 

The  actual  dimensions  of  the  material  are  to  be  used  in  designing 
and  the  various  parts  of  the  forms  are  to  be  made  from  the  following 
stock : 


Actual  Size  Nominal  Size 

Lagging      for  i|*by  3f,  5!  "or  7! "    I'or  if  by  4',  6',  or  8" 

Planks         if"  by  5  f ,  7}  or  gf  2 "  by  6",  8"  or  10" 

Battens       ij"  by  3!"  rough  2  "by  4* 

Joists          if'byaKsl.or?!  2"by4*,6",or8" 

Yokes         1 1  "by  3!  "rough  2  "by  4" 

Posts  3 1  "by  3!"  rough  4"  by  4" 

Caps  3! "  by  3!"  rough  4"  by  4" 

Braces        J"  by  35*  rough  i"by4" 

Stringers     3J'by  sf,  sJ'by  sf'  4' by  6",  6*by6',or6'by  8', 
or  sf  "by  7  J"  rough 


DESIGN    OF   THE    FORMS 

Span  of  floor  beams 2o'-o" 

Center  to  center  floor  beams p'-o* 

Span  of  girder iS'-o* 

Thickness  of  slab 4* 

Cross-section  of  beam  below  slab i7i*x  7!" 

Cross-section  of  girder  below  slab    29!' x  13?" 

Distance  floor  to  floor    i2/-o* 

Cross-section  of  column 26"  x  26* 

Spacing  of  Joists.  The  weight  of  slab  and  construction  live  load 
is  125  pounds  per  square  foot  or  10.4  pounds  per  linear  inch  of  12 
inch  strip.  The  Iofai2"xJ"  section  is  .67  biquadratic  inches  and 
the  safe  span  of  the  lagging  by  Equation  4  is  3 1  .o  inches. 

When  the  span  of  the  joist  is  more  than  about  5  feet,  2"  x  8" 
joists  are  requirecTin  order  that  they  may  be  spaced  far  enough 
apart  to  develop  a  reasonable  percentage  of  the  strength  of  the 
lagging.  It  then  becomes  more  economical  to  support  the  joists 
midway  between  floor  beams,  allowing  the  use  of  much  smaller 
joists. 

The  distance  between  the  centers  of  the  joist  bearers  attached  to 
adjacent  floor  beams  is  108  -  7!  -  2  (i|  +  if )  -  2  =  92  J  inches. 
Then  with  a  center  support  the  span  of  the  joist  is  about  46  inches. 

-  of  if*  x  5!"  joist 10.35  cubic  inches 

|  of  if  x3f*  joist 4.40     " 

By  Equation  7  W  =  2700  pounds  for  i|r  x  5  J"  joist 
and          1 150  pounds  for  i|  *  x  3! *  joist 


32 

The  required  spacing  of  the  larger  joist  is  — —  =  67.6  inches 

12 

and  of  the  smaller  28.8  inches.     The  latter  is  very  close  to  the 
spacing  allowed  by  the  lagging  and  the  smaller  joist  is  adopted. 
The  actual  spacing  of  the  joists  may  be  determined  as  follows : 

Center  to  center  of  girders 20'  -  o" 

Width  of  girder 13  J* 

Side  lagging  2  x  if  =   2  J" 

Battens  2  x  if $\" 

Joist if" 

21  inches 
Center  to  center  of  joists  attached  to  girders 18'  -  3  * 

Approximate  number  of  spaces  — r-f-  7.61 

2o.o 

18'- V 
Therefore,  joists  are  spaced  — — — .  .  .  .27!"  center  to  center. 

o 

Supports  for  the  beam  forms: 

Construction  live  load  per  lineal  foot  of  beam  —  x  7  5  =  48  pounds 

Weight  of  beam  per  linear  foot  — — —  = 174  pounds 

144 

222  pounds 

Total  load  per  linear  inch 18.5  pounds 

I  of  7  f  "x  if"  plank 3.46  pounds 

Safe  span  by  Equation  4 44.2    inches 

Posts  supporting  the  beam  form: 

Construction  live  load  per  linear  foor  I  -     -^—  I  7  5  =  388  pounds 

Weight  of  slab  per  linear  foot  /  -  -  J  50  = 258  pounds 

Weight  of  beam  below  slab  per  linear  foot 

I7^X7JXI5°  =   .  .141  pounds 

144 


787  pounds 

Total  load  per  linear  inch 65.6  pounds 

Safe  bearing  of  post  on  fibres  of  cap  3jx3jx3oo=   4500  pounds 


33 

% 

Safe  spacing  of  posts  j~-r    —  . ; ;;..•..: 68.6  inches 

The  posts  are  spaced  54!  inches  center  to  center,  thus  bringing 
a  post  under  each  alternate  joist.  Under  the  remaining  joists  a 
plank  support  for  the  beam  form  is  placed. 

Center  support  for  joists  (Stringer): 

The  load  sustained  by  each  joist  over  a  span  of  46  inches  is : 

Construction  live  load  -^-      -  x  75  = 656  pounds 

144 

Slab  load  -"•   -  x  50  = 437  pounds 


1093  pounds 

One  post  under  the  center  of  the  stringer  would  sustain  the  load 
from  the  joist  without  injuring  the  fibres  of  the  stringer,  but  it 
would  require  a  6*  x  8"  stringer.  By  placing  a  post  under  each 
alternate  joist  the  bending  moment  for  the  span  of  54!  inches  is 

I  — —  x  27!  I  T8^  =  i  igyoinch  pounds.     Then  bd2  =  59.9  and  with 

b  =  3!  inches,  d  =  3.93  or  5!  inches.  By  using  a  6"  x  4'  stringer 
with  three  posts  the  amount  of  lumber  is  less  than  with  a  6"  x  8* 
stringer  and  only  one  post  and  the  use  of  the  heavier  timber  is 
avoided. 

Supports  for  the  girder  forms.  In  a  manner  similar  to  that  used 
in  determining  the  spacing  of  supports  for  the  beam  forms  the 
safe  spacing  is  found  to  be  39.3  inches. 

Posts  supporting  the  girder  forms.  The  joist  adjacent  to  the 
girder  is  supported  by  the  girder  forms  which  are  in  turn  supported 
by  the  posts. 

Area  of  slab  supported  per  linear  foot  12'  (21"  +  27!*) 

Load  from  slab  per  linear  foot —  xi25=   ..     504  pounds 

144 

Weight  of  girder  below  slab  per  linear  foot 

— —  x  150  =   412  pounds 

916  pounds 
Total  load  per  linear  inch 76.3  pounds 


34 

• 

Safe  spacing  of  posts  ...........................  59.0  inches 

Reaction  from  stringer  not  included  in  the  above,  547  pounds. 
The  safe  distance  between  the  center  and  adjacent  posts  is 

2     =55.4  inches 
76-3 

The  actual  spacing  of  the  posts  may  be  determined  as  follows  : 
Center  to  center  of  column  .......................      18'  -  or 

Width  of  column    ..............................      26" 

Side  lagging  2  x  i|  ..............................      2  J 


Clear  distance  between  column  forms  ..............     15'—  7!  ' 

With  one  post  in  the  center  and  one  on  each  side  47  inches  (46  Tf  ) 
center  to  center  the  full  distance  will  be  covered.  A  plank  support 
may  be  placed  midway  between  the  posts,  but  as  the  clear  span 
between  the  posts  is  only  43  inches  this  is  hardly  necessary. 

Column  Forms: 

lof  i2"x  i|"section  ..........................      1.42 

Pressure  at  bottom  on  strip  1  2  "  x  i  "  =  -        -....    80  pounds 

Pressure  at  J  point  on  strip  1  2  "  x  i  "  =  ...........   60  pounds 

Pressure  at  \  point  on  strip  1  2  "  x  i  "  .............   40  pounds 

Pressure  at  f  point  on  strip  1  2  "  x  i  "  .............    20  pounds 

SAFE    SPAN   OF    LAGGING 

Point  By  Equation  4  By  Equation  8 

B  ott  om  20.2  inches  19.4  inches 

22.2  inches  22.4  inches 

25.4  inches  27.5  inches 

32.0  inches  38.9  inches 

The  span  of  the  2  x  4  is  about  26  +  2  (if  +  i  J  -f  i)  =  34  inches 
lof  2  X4  ....................................   9.1 

W  by  Equation  5    ............................    1325  pounds 

SAFE    SPACING    OF    YOKES 


Bottom  Qgo  7.  6  inches 

oO  X  y^ 

J  10.2  inches 

\  1  5.  3  inches 

30.5  inches 


35 

As  the  strength  of  the  yokes  governs  their  spacing,  the  bolts 

1^2^ 

must  take  a  stress  of  J  —  pounds  requiring  a  section  of  .0414  square 
inches.     A  f  inch  bolt  has  a  net  area  of  .068  square  inches  and  is 


adopted.     A  washer  having  a  net  area  of  —  —  —  =2.21  square 

2  x  300 

inches  and  a  gross  area  of  2.40  square  inches  must  be  used.  A  if 
inch  round  washer  furnishes  this  area  and  a  thicknes  of  |  inch  is 
sufficient. 


36 


DESIGN    OF   A    COMBINED    FOOTING 

When,  on  account  of  limited  ground  area  it  is  impossible  to 
construct  simple  spread  footings  to  support  the  exterior  columns, 


.  K^fcfer-r-K 


[ 


i 


200000                                                       F?= 
71     ri"$ STIRRUPS- KIN  EACH  ROW               r.~^JTJ, 
I     K x-»-r» l"_ 


i"$  STIRRUPS 


I    I..L 


7J 

U 


ABOUT  eO"C.TOC 

FIGURE  10 


either  a  cantilever  construction  must  be  used  or  the  footing  must 

be  combined  with  one  or  more  of  those  for  the  interior  columns. 

The  simplest  type  of  a  combined  footing  is  a  trapezoidal  slab 

the  center  of  gravity  of  which  coincides  with  the  center  of  gravity 


37 

of  the  loads  that  the  footing  sustains.  The  following  design  shows 
the  general  manner  of  determining  the  shape  of  this  type  of  footing, 
its  depth,  and  the  necessary  amount  of  reinforcement. 

Two  columns  i2/-o*  center  to  center,  one  having  a  cross-section 
of  1 8"  x  1 8"  and  the  other  a  cross-section  of  24"  x  24*  sustain  loads 
of  200000  pounds  and  300000  pounds  respectively.  The  allowable 
pressure  on  the  soil  is  5000  pounds  per  square  foot,  and  the  allow- 
able bond  stress  on  the  reinforcing  rods  is  100  pounds  per  square 
inch.  Otherwise  the  specifications  heretofore  used  are  to  be  fol- 
lowed. (See  Figure  i  o). 

The  bearing  area  required  for  these  loads  is  100  square  feet. 
Allowing  the  footing  to  project  six  inches  beyond  the  edge  of  the 
larger  column  the  total  length  of  the  footing  is  14.25  feet. 

The  center  of  gravity  of  the  footing  must  be  at  a  distance  of 

-xi2  +  1.5  =  6.30  feet  from  b\.     Using  the  equations  involv- 
500000 

ing  the  area  and  center  of  gravity  of  a  trapezoid 

,    -  2  X  100 

bi  +  t>2  =   -        —  =  14-04 
14-25 

and 

bl  +  2b2^3X  14.04x6.30  = 

14-25 

whence  bi  =  9.44  feet  and  b2  =  4.60  feet.  The  width  of  the  foot- 
ing at  a  point  midway  between  the  two  columns  is  9.44  - 

14.25 

(9.44  -  4.60)  =  6.94  feet  and  the  center  of  gravity  of  the  larger  of  the 

two  trapezoids  on  either  side  of  this  section  is  —  I  -~ ; ^^  1 

3   y    6.94+9-44     / 

=  3.94  feet  from  this  mid-point.  The  bending  moment  at  this 
section  is  300000  (6.0-3.94)12  =  7,310,000  inch  pounds  and  the 
approximate  depth  of  footing  required  is  28.6  inches.  Allowing 
for  insulation,  the  total  depth  of  the  footing  is  assumed  as  3  2  inches, 
which  adds  to  the  weight  to  be  sustained  400  pounds  per  square 
foot. 

The  above  computation  was  only  for  the  purpose  of  determining 
the  approximate  weight  of  the  footing.  Final  dimensions  may 
now  be  computed  in  the  same  manner. 

The  area  required  is  108.7  square  feet,  and  the  revised  values  of 
bi  and  02  are  10.28  feet  and  4.98  feet  respectively.  The  section  of 


true  maximum  moment  is  of  course  at  the  point  where  the  shear  is 
equal  to  zero.  Let  the  distance  of  this  point  from  bi  be  called  y, 
the  load  on  the  larger  column  PI,  and  the  difference  between  the 
allowable  soil  pressure  and  the  weight  of  the  footing  w.  Then  PI 

=  w  I    l"     —  —  r-^-       1  ,  from  which  the  distance  y  is  determined 

V  / 

as  7.33  feet.  The  width  of  the  footing  at  this  point  (64)  is  7.55  feet 
and  the  distance  of  the  center  of  gravity  of  the  trapezoid  bounded 
by  61  and  64  is  3  .85  feet  from  64. 

The  maximum  bending  moment  is  300000   (7.33  -  1.5)  -  3.85 

I  •  1  7.33x4600  =  590000  foot  pounds  or  7080000  inch 

pounds  and  the  depth  of  footing  required  is  27.0  inches. 

The  width  of  the  transverse  distributing  beam  under  the  larger 
column  is  taken  as  36  inches  and  that  under  the  smaller  column  as 
1  8  inches.  The  moment  at  the  edge  of  the  column  due  to  the 

upward  pressure  of  the  soil  is  f-      —  (10.28-  2.o)2  x  12  =  3000000 

oXIO.2o 

inch  pounds  for  the  longer  beam,  and  -          -  (4.98-1.5)2  x  12  = 

o  x  4-9^ 

732000  inch  pounds  for  the  shorter  beam.     The  maximum  shear  at 

the  edge  of  the  column  for  the  former  is  -  --  1  -  1   = 

2       y      10.28      j 

1  2  1000  pounds,  and  for  the  latter  70000  pounds. 

The  depths  required  are  27.9  inches  and  19.5  inches  respectively. 
A  depth  of  28  inches  is  adopted  for  the  entire  footing  and  with  four 
inches  of  insulation  the  total  depth  is  32  inches  as  assumed. 

The  distance  from  the  point  of  zero  shear  to  the  point  where  the 
unit  shear  reaches  the  allowable  value  for  plain  concrete  is  deter- 


mined by  solving  the  following  equation  for  z.     z2 

-+-  zwb4-  —  -  —   -          -v'jdb4  =  o    (In  using  this  equation  inches 


I  —  .  —  (bi-b2)  I 


must  be  used  with  pounds  per  square  inch,  and  feet  with  pounds 
per  square  foot).  From  the  above  equation  the  value  of  z  is  2.70 
feet,  and  the  width  of  the  footing  at  this  point  (66)  is  8.56  feet  and 
that  at  the  edge  of  the  larger  column  (66)  is  9.3  5  feet.  The  distance 
over  which  web  reinforcement  is  required  (q)  is  2.13  feet.  The 


39 
maximum  shear  along  66  is^-^  -  ^^  (4-83)   (4600)    =  187700 

2 

pounds,  and  the  unit  shear  68.3  pounds.  The  average  unit  shear 
over  the  distance  q  is  54.  1  5  pounds,  the  average  width  of  the  footing 
8.95  feet  and  the  total  area  of  web  reinforcement  required 

2  x  ^4.iqx8.o';x2.i^x  144  ,  .  .      A,  .  L 

-  —  6.20  square  inches,  requiring  thirty- 
3  x  16000 

two  \  inch  round  single  stirrups.  The  spacing  of  the  stirrups  is 
determined  graphically  on  Figure  10. 

In  a  similar  manner  forty-two  \  inch  round  single  stirrups  spaced 

over  a  distance  of  3.01  feet  are  required  at  the  other  end  of  the 

footing.    Computing  the  horizontal  reinforcement  for  the  main  slab 

7080000 


16000  x.  875x28 


.     ,  , 

=  1  8.  i  square  inches  and 


20  =    -          °  —  -  =  76.6  inches;   which  will  be  furnished  by 
IQOX  .875x28 

twenty-five  i  inch  round  rods.  Similarly  for  the  transverse  beams 
sixteen  rods  and  nine  rods  respectively  are  required. 

The  same  condition  with  regard  to  diagonal  tension  failures  may 
be  assumed  for  the  distributing  beams  as  was  assumed  in  the  square 
footing. 

Computing  the  unit  shear  28  inches  from  the  edge  of  the  column 
it  is  found  to  be  59.7  pounds  so  that  stirrups  are  required  from  tjiis 
point  to  the  point  where  the  shear  is  equal  to  40  pounds  per  square 
inch  which  is  at  a  distance  of  33  inches  from  the  edge  of  the  column. 

2  x  40.  o  x  36  x  5 

The  area  of  web  reinforcement  required  is  -  f^  =  .374 

3  x  16000 

square  inches  so  that  one  double  \  inch  round  stirrup  will  be  suffi- 
cient. Two  double  stirrups  are  placed  30  inches  from  the  edge  of 
the  column  in  order  to  distribute  the  reinforcement  over  the  full 
width  of  the  beam. 


Gaylord  Bros. 

Maker* 

Syracuse  N.  Y. 
FAT.  JM.  2 1,1  Ml 


YC   13071 


f&i 


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